Re: Question: RT schedular : task_tick_rt(struct rq *rq, struct task_struct *p) : decreases overhead when rq->nr_running == 1

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On 08/08/07, Ingo Molnar <[email protected]> wrote:
> * Mitchell Erblich <[email protected]> wrote:
> > After
> >     p->time_slice = static_prio_timeslice(p->static_prio);
> >
> > Why isn't their a check like
> >     if  (rq->nr_running == 1)
> >          return;
> >
> > Which world remove the need for any recheduling or requeue'ing...
> your change is a possible optimization, but this is a pretty rare
> codepath because the overwhelming majority of RT apps uses SCHED_FIFO.
> Plus, the time_slice going down to 0 is a relatively rare event even for
> SCHED_RR tasks. [ ... ]

I doubt it's a worth optimization. It's a rare codepath and, moreover,
this optimization is sub-optimal. e.g. the rest of tasks that
contributed to 'rq->nr_running > 1' could be SCHED_NORMAL (or of lower
rt_prio) and so resched_task() is useless as well.

I guess, the following thing would do a better job:

- we do need reschedule() _only_  if there are other task on the
_same_ queue (for this prio level) :

(1) if there is a task with a higher prio, resched_task() would be
called in other place;
(2) if there are tasks of lower prio, we don't care.

I may be wrong, but at the first glance it looks like the following
trick would do the job.. no?

p.s. even if it's ok, it's still a rare codepath (although, it adds
some code to the rare path.. so the drawback is only code-size, I

--- kernel/sched_rt-prev.c      2007-08-09 09:55:10.000000000 +0200
+++ kernel/sched_rt.c   2007-08-09 09:58:53.000000000 +0200
@@ -207,6 +207,11 @@ static void task_tick_rt(struct rq *rq,

        p->time_slice = static_prio_timeslice(p->static_prio);
+       /* We are the only element on the queue. */
+       if (p->run_list.prev == p->
+               return;

        /* put it at the end of the queue: */

Best regards,
Dmitry Adamushko
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